TESTING THE EQUALITY OF TWO VARIANCES THE F

TESTING THE EQUALITY OF TWO VARIANCES: THE F TEST Application test assumption of equal variances that was made in using the t-test p interest in actually comparing the variance of two populations p fdist 1

The F-Distribution Assume we repeatedly select a random sample of size n from two normal populations. p Consider the distribution of the ratio of two variances: F = s 12/s 12. p The distribution formed in this manner approximates an F distribution with the following degrees of freedom: v 1 = n 1 - 1 and v 1 = n 1 - 1 p fdist 2

Assumptions Random, independent samples from 2 normal populations p Variability p fdist 3

F-Table The F table can be found on the appendix of our text. It gives the critical values of the Fdistribution which depend upon the degrees of freedom. fdist 4

Example 1 p Assume that we have two samples with: n 2 = 7 df = 7 -1= 6 p and n 1 =10 df = 10 -1= 9 Let v = F(6, 9) where 6 is the df from the numerator and 9 is the df of the denominator. p Using the table with the appropriate df, we find : P(v < 3. 37) = 0. 95. fdist 5

Example 2: Hypothesis Test to Compare Two Variances 1. Formulate the null and alternate hypotheses. H 0: s 1 2 = s 1 2 > s 2 2 Ha: [Note that we might also use s 12 < s 22 or s 12 =/ s 22] 2. Calculate the F ratio. F = s 12/s 12 [where s 1 is the largest or the two variances] 3. Reject the null hypothesis of equal population variances if F(v 1 -1, v 2 -1) > Fa [or Fa/2 in the case of a two tailed test] fdist 6

Example 2 The variability in the amount of impurities present in a batch of chemicals used for a particular process depends on the length of time that the process is in operation. Suppose a sample of size 25 is drawn from the normal process which is to be compared to a sample of a new process that has been developed to reduce the variability of impurities. n s 2 Sample 1 25 1. 04 Sample 2 25 0. 51 fdist 7

Example 2 continued H 0: Ha: s 1 2 = s 2 2 s 1 2 > s 2 2 F(24, 24) = s 12/s 22 = 1. 04/. 51 = 2. 04 Assuming a = 0. 05 cv = 1. 98 < 2. 04 Thus, reject H 0 and conclude that the variability in the new process (Sample 2) is less than the variability in the original process. fdist 8

Try This A manufacturer wishes to determine whethere is less variability in the silver plating done by Company 1 than that done by Company 2. Independent random samples yield the following results. Do the populations have different variances? [solution: reject H 0 since 3. 14 > 2. 82] fdist 9